SOLUTION: The function f(x)=x^3-x+1 has exactly one real zero. It is between (A) -2 and -1 (B) -1 and 0 (C) 0 and 1 (D) 1 and 2 (E) 2 and 3 Can you please tell me how to get this

Question 755154: The function f(x)=x^3-x+1 has exactly one real zero. It is between
(A) -2 and -1
(B) -1 and 0
(C) 0 and 1
(D) 1 and 2
(E) 2 and 3
Can you please tell me how to get this answer? its on my final and i don't remember ever doing them...
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
More sense made by looking at because this is 1 unit less than f(x). g(x) here is factorable and easily analyzed for its intervals regarding signs of the values for graphing of g(x). You could then look at a graphical representation and visually judge WHERE when the graph is raised 1 unit, can exactly 1 point intersect the x axis.

If raising the graph until the point between 0 and 1 is reached, then there still is one other point way on the left which also intersects the x axis. Keep raising the graph until f(x) is reached, one unit above g(x). The intersecting point is way over on the left. This is at .

and
and find that

The intervals around the roots for g(x) are
(-infin,-1)
(-1, 0)
(0,+1)
(+1, +infin)
Check the signs of those intervals and show this on a graph, even if only qualitatively between the within each interval.
Also notice qualitatively that g(x) and therefore f(x) decrease as x tends toward lesser than -1 while g(x) has a root at x=-1.

Reminding again that the one real zero is at a particular point for , probably your choice (A) for -2 to -1.
f(x)

g(x)

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The function f(x)=x^3-x+1 has exactly one real zero. It is between
(A) -2 and -1
(B) -1 and 0
(C) 0 and 1
(D) 1 and 2
(E) 2 and 3
Can you please tell me how to get this answer? its on my final and i don't remember ever doing them..
***
There is a rule I found called the Location Theorem which can be applied here.
This is how it works: plug in consecutive x-values like a and b.
If f(a) and f(b) have opposite signs, there is at least one real root between a and b.
..
(A) a=-2, b=-1
f(a)=f(-2)=-8+2+1=-5
f(b)=f(-1)=-1+1+1=1
f(a) and f(b) have opposite signs, so there is a real root between -2 and -1
..
(B) a=-1, b=0
f(a)=f(-1)=1
f(b)=f(0)= 1
f(a) and f(b) have same signs, so there isno real root between -1 and 0
..
(C) a=0, b=1
f(a)=f(0)=1
f(b)=f(1)=1
f(a) and f(b) have same signs, so there is no real roots between 0 and 1
..
You can try (D) and (E) yourself.
If you have a graphing calculator, you will find a real zero at -1.32472.., confirming that A is the ans.
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